## #42 In Post-Module Assessment [ENDORSED]

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

Mario Reyes 1C
Posts: 45
Joined: Fri Apr 06, 2018 11:04 am

### #42 In Post-Module Assessment

The question asks 'An excited hydrogen atom emits light with a frequency of 1.14 x 10^14 Hz to reach the energy level n = 4. In what principle quantum level did the electron begin?'
A. n = 5

B. n = 6

C. n = 4

D. n = 7

I am not sure how to approach this problem. Would we have to plug in every quantum level to find delta E and see which one comes out with a frequency of 1.14x10^14 Hz, or is there an easier way to solve this?

NatalieSDis1A
Posts: 39
Joined: Fri Apr 06, 2018 11:05 am

### Re: #42 In Post-Module Assessment

I am also stuck on this one but I do not think you plug in every quantum level. I began by using the formula En = (-hR/n2). If we consider 1.14 X 1014 to be the delta E, I think the formula delta E = e final - E initial should work. My progress so far looks like this:

1.14 X 1014 = E4 - Ex
1.14 X 1014 = (-hR/16) - (-hR/x2)
Ex = -2.11 X 10-19 = (-hR/x2)
x2 = 10.318
x = n = 3.2

This doesn't make sense because it is not a whole number. Does anyone know what I'm doing wrong.

NatalieSDis1A
Posts: 39
Joined: Fri Apr 06, 2018 11:05 am

### Re: #42 In Post-Module Assessment

Update: I figure it out!

I made the mistakes of not converting the frequency to energy and subtracting in the wrong order.

E = hv

E = h (1.14 X 10^14) = 7.55 X 10^-20

7.55 X 10^-20 = Ex - E4

Ex = (-hr)/x^2 = -6.06 X 10^-20

x^2 = 35.9

x = n = 6

Mario Reyes 1C
Posts: 45
Joined: Fri Apr 06, 2018 11:04 am

### Re: #42 In Post-Module Assessment

Since it is Efinal- Einitial, shouldn't it be E4-Ex?

Also, I am confused on these steps. How did you get -6.06x10^-20?

Ex = (-hr)/x^2 = -6.06 X 10^-20

x^2 = 35.9

x = n = 6

NatalieSDis1A
Posts: 39
Joined: Fri Apr 06, 2018 11:05 am

### Re: #42 In Post-Module Assessment  [ENDORSED]

You're right it should be E4 - Ex. Since, I used a positive energy I was able to switch them. Really, the 7.55 X 10^-20 should be negative since we are losing energy. If you don't switch them is look like this:

-7.55 X 10^-20 = E4 - Ex

We also can calculate that E4 = (-hr/16) = -1.36 X 10^-19

Using algebra we add 7.5 X 10^-20 to each side and move Ex to the opposite side. That process gives you Ex = -6.06 X 10^-20.

We know that Ex = -hr/x^2 so we plug in the Ex value we just calculated leaving us with a solvable equation.

abbydouglas1K
Posts: 65
Joined: Fri Sep 28, 2018 12:26 am

### Re: #42 In Post-Module Assessment

This was super helpful thank you!